product of non-empty set of non-empty sets is non-empty

Proposition 1.

Proof.

Suppose C={Aj∣j∈J} is a set of non-empty sets, with J≠∅. We want to show that

B:=∏j∈JAj

is non-empty. Let A=⋃C. Then, by AC, there is a functionf:C→A such that f⁢(X)∈X for every X∈C. Define g:J→A by g⁢(j):=f⁢(Aj). Then g∈B as a result, B is non-empty.
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Remark. The statement that if J≠∅, then B≠∅ implies Aj≠∅ does not require AC: if B is non-empty, then there is a function g:J→A, and, as J≠∅, g≠∅, which means A≠∅, or that Aj≠∅ for some j∈J.

Proposition 2.

(*) implies AC.

Proof.

Suppose C is a set of non-empty sets. If C itself is empty, then the choice function is the empty set. So suppose that C is non-empty. We want to find a (choice) function f:C→⋃C, such that f⁢(x)∈x for every x∈C. Index elements of C by C itself: Ax:=x for each x∈C. So Ax≠∅ by assumption. Hence, by (*), the (non-empty) cartesian productB of the Ax is non-empty. But an element of B is just a function f whose domain is C and whose codomain is the union of the Ax, or ⋃C, such that f⁢(Ax)∈Ax, which is precisely f⁢(x)∈x.
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